1023 Have Fun with Numbers (20分)

Notice that the number 123456789 is a 9-digit number consisting exactly the numbers from 1 to 9, with no duplication. Double it we will obtain 246913578, which happens to be another 9-digit number consisting exactly the numbers from 1 to 9, only in a different permutation. Check to see the result if we double it again!

Now you are suppose to check if there are more numbers with this property. That is, double a given number with k digits, you are to tell if the resulting number consists of only a permutation of the digits in the original number.

Input Specification:

Each input contains one test case. Each case contains one positive integer with no more than 20 digits.

Output Specification:

For each test case, first print in a line "Yes" if doubling the input number gives a number that consists of only a permutation of the digits in the original number, or "No" if not. Then in the next line, print the doubled number.

Sample Input:

1234567899
Sample Output:

Yes
2469135798

思路

高精度 X 低精度

#include<iostream>
#include<string>
#include<vector>
#include<algorithm>

using namespace std;

const int N = 30;

vector<int> a;
int st[N][2];

int main(){
    string p;
    
    cin >> p;
    
    for(int i = p.size() - 1; i >= 0; i --) a.push_back(p[i] - '0'), st[p[i] - '0'][0] ++;
    
    int x = 0;
    for(int i = 0; i < a.size(); i ++){
        a[i] = a[i] * 2 + x;
        x = a[i] / 10;
        a[i] %= 10;
    }
    
    if(x) a.push_back(x);
    reverse(a.begin(), a.end());

    int flag = 0;
    
    for(auto t : a) st[t][1] ++;
    
    for(int i = 0; i < 10; i ++)
        if(st[i][1] != st[i][0]){
            flag = 1;
            break;
        }
    
    if(flag) puts("No");
    else puts("Yes");
    for(auto t : a) cout << t;
    return 0;
}